Exact value of the correlation factor for the divacancy ... - arXiv

Exact value of the correlation factor for the divacancy mechanism in FCC crystals: the end of a long quest J.L. Bocquet CMLA, LRC-Méso, ENS CACHAN, 61 Avenue du Président Wilson, 94235 Cachan, France [email protected]

Introduction Matter transport in crystals is most often mediated through point defects and atomic species are not random walkers. Their walk departs from that of a truly random walk by a factor, generally smaller than unity, called the correlation factor. After a first initial jump has been performed by an atom, a simple inspection of the mechanism at the atomic scale shows that the direction of its next jump has a larger than random probability to happen in the reverse direction. The net result is a decrease in efficiency of the walk.1 Taking into account the effect of the immediate reverse jump only gives a good order of magnitude of the correlation effect. However the exact value requires a closer examination of the atomic mechanism. The most simple case ever studied is the diffusion of a tagged atom called ‘the tracer’ moved by a single defect (vacancy or interstitial) in an infinite crystal lattice. The chemical nature of the tracer can be identical (‘self diffusion’) or different (‘solute diffusion’) from that of all other atoms. In both cases the calculation of the correlation

 effect boils down to evaluate the probabilities pω1 ,ω2 of performing a tracer jump ω2

 following a previous jump ω1 , with the help of the same defect. The average quadratic displacement of the tracer due to its encounter with a given defect is a function of the number of jumps and of the probabilities pω1 ,ω2 . Due to the translational symmetry of the medium, the atomic configuration after a tracer jump is identical to that before the jump, apart from a rotation: the probabilities pω1 ,ω2 do not depend on







 ω1 and ω2 separately, but only on ω2 − ω1 . In such an ideal situation, a simple recurrence equation for the displacement of the tracer atom can be written and solved in order to extract the mean quadratic displacement of the tracer due to its encounter with a unique defect. Further, choosing the x-axis as a principal axis of the diffusion tensor restrains the problem to the only consideration of the x-components x1 and x2 . 2 For the monovacancy mechanism, all the jumps have the same x-component ‘s’ and only one type of jump has to be defined. The correlation factor fV is a simple function of the average product t1 =< x1 x2 > / s 2 = p + − p − where p + , p − are the probabilities of performing two successive tracer jumps with an x-component of the same sign or not. After the first jump, the defect is in a convenient position to let the tracer atom perform immediately a reverse jump, which explains that p − > p + , and t1 < 0 . The correlation factor is given by: fV = (1 + t1 ) / (1 − t1 ) . For the more complicated divacancy mechanism, all the jumps have the same x-component but two types of jumps must be defined according to the tracerdivacancy configuration. The correlation factor f 2V is a function of the four average

products tαβ =< x1α x2β > , where the superscripts α,β stand for the type of the jump ( α,β = 1, 2 ). The two types of jump are defined, according to the following rule: * a type 1 jump is performed if the two vacancies are located on the same x plane before the jump is completed; * a type 2 jump is performed if only one vacancy is located on the same x plane as that of the tracer before the jump is completed. In this case, the correlation factor is given by: (1 + t21 )(1 + t12 ) − t11t22 f 2V = (1) (1 − t11 )(1 − t22 ) − t12t21 Locating the origin of the coordinates on the site occupied by the tracer atom after its

 first jump ω1α (x-component x1α ) , the problem consists in calculating the total

 probability for the defect, starting initially at − ω1α to come back in the neighbourhood



 of the tracer atom at − ωβ2 and make it perform a further jump ωβ2 (x-component x2β ). The precision of the result is a (slowly) varying function of the number of jumps, the defect is allowed to perform before its next exchange with the tracer: * the first evaluation, made by hand-summing the contributions of short defect trajectories, evaluated the divacancy correlation factor to 0.54; 3 * the second evaluation amounting to 0.475 rested on a matrix method and took account of divacancy trajectories of arbitrary length, provided they were confined in a small volume V around the tracer atom;4 * the third evaluation, using an extended version of the matrix method (i.e. a larger V) decreased the value down to 0.472;5 * the best values known up to now were later provided by direct Monte-Carlo simulations: 0.458 ± 0.001 6 and 0.4582 ± 0.0005. 7 The object of the present contribution is to determine the first exact evaluation of this quantity with an integral method already illustrated in previous publications.8-10 The first part is devoted to notations and definitions. The second part establishes the transport equations for the divacancy mechanism. The third part solves the linear system yielding the average products tαβ together with the final exact value of the correlation factor f 2V . Details of calculations are passed in Appendices: *Appendix A gives useful expressions of the determinant and cofactors entering the solution; * Appendix B determines the coefficients of the linear system yielding the return probabilities of the vacancy on sites giving rise to a next tracer jump; * Appendices C and D establish the formal expressions of coefficients entering the definition of tαβ ; * Appendix E establishes the equality t11 = t22 . 1. Definitions and notations The sites in the host FCC lattice are denoted by: Ω ix+ jy + kz = i Ω x + j Ω y + k Ω z where

i, j, k are integers complying with the condition i+j+k even and the basis vectors are a a a Ω y = [0,1, 0] Ω z = [0, 0,1] , with « a » defined according to Ω x = [1, 0, 0] 2 2 2 standing for the FCC lattice parameter. The centre of the divacancy defines unambiguously the position of the two vacancies: it moves on a new lattice which is made up by the middles of all nearest neighbour bonds of the parent FCC lattice. The new sites will be described with the help of new basis vectors defining the new unit length along the three coordinate a a a ω y = [0,1, 0], ωz = [0, 0,1] . The orientation axis of the axis: ωx = [1, 0, 0], 4 4 4 divacancy is given by the direction of the line segment joining the two vacancies of the pair. Three different types of sites are to be considered and denoted with a subscript ‘x’ , ‘y’ or ‘z’, the latter being related to the plane containing the two ends of the divacancy: * ‘x’ sites correspond to divacancies contained in planes x = 2i ; the ‘x’ sites are located at r2ix +(2 j ±1) y + (2 k ±1) z = 2i ωx + (2 j ± 1) ω y + (2k ± 1) ωz . The orientation of the divacancy is a function of the parity of ‘i’. If ‘i’ is even, the orientation of the divacancy is oriented along ω y + z for sites r2ix +(2 j +1) y +(2k +1) z and r2ix +(2 j −1) y +(2k −1) z , and oriented along ω y − z for sites r2ix +(2 j +1) y +(2k −1) z and r2ix +(2 j −1) y +(2k +1) z . The reverse is true when ‘i’ is odd. The first neighbours of ‘x’ sites belonging to plane x = 2i lie only in adjacent x = 2i + 1 and x = 2i − 1 planes and are exclusively of ‘y’ or ‘z’ type; * ‘y’ sites correspond to divacancies contained in plane y = 2 j ; the ‘y’ sites are located at r(2i ±1) x+ 2 jy +(2 k ±1) z = (2i ± 1) ωx + 2 j ω y + (2k ± 1) ωz ; the orientation of the divacancy follows similar rules as above after a cyclic permutation i → j , j → k , k → i but now depends on the parity of ‘j’. If ‘j’ is even, the orientation of the divacancy is oriented along ωz + x for sites r(2i +1) x +2 jy +(2 k +1) z and r(2i −1) x + 2 jy +(2k −1) z , and oriented along ωz − x for sites r(2i +1) x + 2 jy +(2 k −1) z and r(2i −1) x + 2 jy +(2k +1) z . The reverse is true when ‘j’ is odd. The first neighbours of ‘y’ sites belonging to plane y = 2 j lie only in adjacent y = 2 j + 1 and y = 2 j − 1 planes and are exclusively of ‘z’ or ‘x’ type; * ‘z’ sites correspond to divacancies contained in plane z = 2k ; the ‘z’ sites are located at r(2i ±1) x +(2 j ±1) y + 2kz = (2i ± 1) ωx + (2 j ± 1) ω y + 2k ωz ; the orientation of the divacancy depends on the parity of ‘k’. If ‘k’ is even, the orientation of the divacancy is oriented along ωx+ y for sites r(2i +1) x +(2 j +1) y + 2kz and r(2i −1) x +(2 j −1) y + 2kz , and oriented along ωx− y for sites r(2i +1) x +(2 j −1) y + 2kz and r(2i −1) x +(2 j +1) y + 2kz . The reverse is true when ‘k’ is odd. The first neighbours of ‘z’ sites belonging to plane z = 2k lie only in adjacent z = 2k + 1 and z = 2k − 1 planes and are exclusively of ‘x’ or ‘y’ type.

Fig. 1. The empty circles denote the sites of the host FCC lattice. The filled circles denote the new lattice sites swept by the divacancy centre. Sites coloured in red represent the sink sites around the origin. The continuous thin lines (red or back) represent the possible first neighbour transitions between new sites. The eight first neighbours of ‘y’ site ω x + z are displayed.

The current vector in the new lattice will mostly be denoted by the symbol ‘r’; ‘ω’ will be used for a small number of sites around the origin, which play a particular role in the problem. A picture showing the position of the new sites with respect to FCC lattice sites is displayed in Fig. 1. Each site of type ‘x’, ‘y’ or ‘z’ is represented by a filled circle attached to a small line segment aligned along the x, y, or z axis respectively. An alternative view of this new lattice, without any reference to the host FCC lattice, is displayed in Fig. 2: it consists of a simple cubic array of octahedrons sharing their vertices with a coordination number equal to 8.

Fig. 2. New lattice of coordinance 8 swept by the divacancy centre.

Since the divacancy is not allowed to dissociate, the two constituting vacancies can only exchange with one of the four first neighbours they have in common on the FCC host lattice; hence a total of eight possible jumps for the migration of the complex. Three density probability functions L(xυ) (r , t ), L(yυ) (r , t ), L(zυ) (r , t ) will be introduced for the probability of finding a vacancy on the three types of site at time t; the superscript (υ) keeps track of the initial condition, i.e. the location of the divacancy after the first tracer jump of type (υ) . A picture of the two types of jump is displayed on Fig. 3.

Fig. 3. The empty circles denote the sites of the host FCC lattice. Squares denote the position of the vacancies. The filled circles denote the sites of the corresponding divacancy centre. The arrows with thick lines denote jumps available for the tracer located on the origin.

The statement of the correlation problem then runs as follows: assuming that the tracer atom just performed a jump of type (υ) with a negative x-component and taking the new tracer site as the origin of the coordinates, the divacancy is modelled by a unit source which is put on its starting site and spreads in time according to diffusion laws. The purpose of the calculation is to determine which fraction of the unit source flows onto each of the sites close to the origin, namely the four ‘x’ sites ω y + z , ω y − z , ω− y + z , ω− y − z of plane x=0, the four ‘y’ sites ωz + x , ωz − x , ω− z + x , ω− z − x of plane y = 0 and the four ‘z’ sites ωx + y , ωx − y , ω− x + y , ω− x − y of plane z = 0 . The arrival of the

vacancy centre on any one of these sites gives rise to a further tracer jump; they play the role of sinks in the calculation of the return probabilities and are characterized by a total escape frequency equal to zero. For instance, the fraction of the unit source flowing onto the sink site ωx + z is given by the time integral of the density probability L(cυ) (ωx + z + ω1V , t ) times the divacancy jump rate WO , where ωx + z + ω1V is a first neighbour of type ‘c’ of ωx + z i.e. ∞

WO ∫ L(cυ) (ωx + z + ω1V , t )dt = WO LL(cυ) (ωx + z + ω1V , p )

p =0

,

0

where LL(cυ) stands for the Laplace transforms of L(cυ) .

(2)

If the divacancy centre is located on sites ωx + z + ω1V = ω2 x + y + z , ω2 x − y + z the type of the next tracer jump is equal to 1; if it is located on sites ωx + y + 2 z , ωx − y + 2 z , the type of the next tracer jump is equal to 2. Summing Eq. 2 on all sink sites gives the total fraction of the unit source flowing back onto the tracer atom. In 3D lattices, this fraction is strictly smaller than unity, since the defect can escape to infinity without any further exchange jump than the first. The isotropy of the matter transport by this mechanism together with the symmetries of the new lattice allow us to use initial conditions (CI) with a four-fold symmetry around the x-axis and a mirror symmetry plane at x=0, as already illustrated in more detail in a previous work devoted to the monovacancy mechanism.10 1.1 Initial condition CI1 A previous type 1 jump has been performed with a negative x-component: this means that, before the jump, the divacancy was in plane x=0 at any of the sites ω y + z , ω y − z , ω− y + z , ω− y − z and the tracer was located at one of the four possible sites ω2 x + 2 y , ω2 x − 2 z , ω2 x + 2 z , ω2 x − 2 y . After completion of this first jump, the number of resulting

possible configurations amounts to eight, with a divacancy located at four ‘y’ sites ωx + 2 y + z , ωx + 2 y − z , ωx − 2 y + z , ωx − 2 y − z or four ‘z’ sites ωx + y + 2 z , ωx + y − 2 z , ωx − y + 2 z , ωx − y − 2 z . The unit source of the diffusion problem is thus replaced by eight sources with a weight +1/8 on the four ‘y’ sites and the four ‘z’ sites of plane x = +1 . The mirror source -1 is replaced by eight sources with weight -1/8 on the mirror-symmetric sites on plane x = −1 , that is the four ‘y’ sites ω− x − 2 y − z , ω− x − 2 y + z , ω− x + 2 y − z , ω− x + 2 y + z and the four ‘z’ sites ω− x − y − 2 z , ω− x − y + 2 z , ω− x + y − 2 z , ω− x + y + 2 z .

In all what follows a shorthand notation will replace cos(lk x + mk y + nk z ) and sin(lk x + mk y + nk z ) by clx + my + nz and slx + my + nz respectively. The indices l , m, n will not be

mentionned when equal to unity. The initial condition for ‘y’ sites and its Fourier transform are then given by: 1  δ(r − ωx + 2 y + z ) + δ(r − ωx + 2 y − z ) + δ(r − ωx − 2 y + z ) + δ(r − ωx − 2 y − z )  L(1) y ( r , t = 0) = 8 1 −  δ(r − ω− x + 2 y + z ) + δ(r − ω− x + 2 y − z ) + δ(r − ω− x − 2 y + z ) + δ(r − ω− x − 2 y − z )  8 (1) → FLy = −ic2 y cz sx

(3.1)

where the superscript refers to the type of the previous jump. The Fourier transform (1) − ikr (1) of L(1) Ly (r , t ) where the summation runs over all y ( r , t ) is defined by FLy ( k , t ) = ∑ e r

sites swept by the centre of the divacancy and kr stands for a scalar product. Due to the four-fold symmetry, the z variable plays the same role as y, hence:

1  δ( r − ωx + 2 y + z ) + δ( r − ωx + 2 y − z ) + δ(r − ωx − 2 y + z ) + δ( r − ωx −2 y − z )  L(1) z ( r , t = 0) = 8 1 −  δ( r − ω− x + 2 y + z ) + δ( r − ω− x + 2 y − z ) + δ( r − ω− x −2 y + z ) + δ(r − ω− x − 2 y − z )  8 (1) → FLz = −ic y c2 z sx

(3.2)

Together with the initial condition for L(1) x (r , t ) : → FL(1) x = 0.

(3.3)

1.2 Initial condition CI2 A previous type 2 jump has been performed with a negative x-component: this means that before the jump the divacancy was located at ωx + 2 y + z ωx + 2 y − z ωx − 2 y + z ωx − 2 y − z or ωx + y + 2 z , ωx + y − 2 z , ωx − y + 2 z , ωx − y − 2 z and the tracer was located at one of the four possible sites ω2 x + 2 y , ω2 x − 2 z , ω2 x + 2 z , ω2 x − 2 y . The resulting configurations amount to four with a tracer atom on the origin and the divacancy located on sites of type ‘x’ at ω2 x + y + z , ω2 x − y + z , ω2 x + y − z , ω2 x − y − z . The unit source +1 of the diffusion problem is thus split into four sources with weight +1/4 on the four ‘x’ sites of plane x = +2 located at ω2 x + y + z , ω2 x − y + z , ω2 x + y − z , ω2 x − y − z ; the mirror source -1 is split into four sources with weight -1/4 on the four ‘x’ sites of plane x = −2 at ω−2 x + y + z , ω−2 x − y + z , ω−2 x + y − z , ω−2 x − y − z . The Fourier transform of the initial condition for the function L(2) x ( r , t ) is given by: (2) x

FL

1 e =  4 

− ik ω2 x+ y+ z

+e

− ik ω2 x+ y − z

+e

− ik ω2 x− y + z

−e

+e

− ik ω−2 x+ y + z

− ik ω2 x− y− z

−e

− ik ω−2 x+ y − z

−e

− ik ω−2 x− y+ z

  − ik ω − e −2 x− y−z 

(4.1)

= −4icx c y cz sx (2) together with the Fourier transformed initial conditions for functions L(2) y ( r , t ), Lz ( r , t ) :

→ FL(2) y =0

(4.2)

→ FL(2) z =0

(4.3)

1.3 Relationships between unknowns due to symmetries The mirror symmetry plane leads to presence probabilities equal to zero on all sites of plane x = 0 , and to presence probabilities which are equal in magnitude but of opposite signs on the sites of planes x = +i and x = −i . The presence probabilities on sink sites, which are first neighbours of the origin, are named with particular subscripts x 0 , y1 , z1 :

L(xυ) (ω y + z , t ) = L(xυ) (ω y − z , t ) = L(xυ) (ω− y + z , t ) = L(xυ) (ω− y − z , t ) = L(xoυ) = 0 L(yυ) (ωx + z , t ) = L(yυ) (ωx − z , t ) = L(yυ1)

L(yυ) (ω− x + z , t ) = L(yυ) (ω− x − z , t ) = − L(yυ1)

L(zυ) (ωx + y , t ) = L(zυ) (ωx − y , t ) = L(yυ1)

L(zυ) (ω− x + y , t ) = L(zυ) (ω− x − y , t ) = − L(zυ1)

(5)

with L(yυ1) = L(zυ1) . In the same way, the ‘y’ and ‘z’ sites in the planes x = +1 and x = −1 which are first neighbours of sinks without being sinks themselves play a particular role in the diffusion problem. They are named with particular subscripts y 2 , z 2 : L(yυ) (ωx + 2 y + z , t ) = L(yυ) (ωx − 2 y + z , t ) = L(yυ) (ωx + 2 y − z , t ) = L(yυ) (ωx − 2 y − z , t ) = L(yυ2) L(yυ) (ω− x + 2 y + z , t ) = L(yυ) (ω− x − 2 y + z , t ) = L(yυ) (ω− x + 2 y − z , t ) = L(yυ) (ω− x − 2 y − z , t ) = − L(yυ2) L(zυ) (ωx + y + 2 z , t ) = L(zυ) (ωx − y + 2 z , t ) = L(zυ) (ωx + y − 2 z , t ) = L(zυ) (ωx − y − 2 z , t ) = L(zυ2)

(6)

L(zυ) (ω− x + y + 2 z , t ) = L(zυ) (ω− x − y + 2 z , t ) = L(zυ) (ω− x + y − 2 z , t ) = L(zυ) (ω− x − y − 2 z , t ) = − L(zυ2)

with L(yυ2) = L(zυ2) . The time integral of the probabilities on sink sites are given by their Laplace transforms calculated with p = 0 ; they are denoted S with the same subscripts and superscripts as above and verify the following relationships: ∞

S

( υ) xo

∫

=

L(xυ) (ω y + z , t )dt = LL(xoυ) ( p )

t =0

p=0

=

1 VZB

∫

FLL(xυ) (k , p )

p=0

e

=

1 VZB

∫

FLL(yυ) (k , p )

p=0

e+ ik ωx+z d3 k

=

1 VZB

∫

FLL(zυ) (k , p )

p =0

e

∞

∫

S y( 1υ) =

L(yυ) (ωx + z , t )dt = LL(yυ1) ( p )

p=0

L(zυ) (ωx + y , t )dt = LL(zυ1) ( p )

p =0

t =0 ∞

S

( υ) z1

=

∫

t =0

+ ik ω y + z

+ ik ωx+ y

d3 k = 0

(7)

d3k

with S y( 1υ) = S z(1υ) because of the four-fold symmetry, VZB the first Brillouin zone and

FLL(xυ) (k , p), FLL(yυ) (k , p), FLL(zυ) (k , p) the Fourier-Laplace transforms of the functions L(xυ) (r , t ), L(yυ) (r , t ), L(zυ) (r , t ) with initial conditions (υ) . Thanks to mirror symmetry: ∞

∫

t =0

∞ ( υ) y

L (ω− x + z , t )dt = − S

( υ) y1

∫

L(zυ) (ω− x + y , t )dt = − S z(1υ)

t =0

The quantity S y( 1υ) = S z(1υ) increases in a monotonic way towards an horizontal asymptote at long times: its Laplace transform will diverge as 1 / p when p → 0 . We will see in Appendix B that S y( 1υ) always appears in the equations as a product pS y( υ1 ) which tends towards a finite value when p → 0 . For the sites in the planes x = +1 and x = −1 which are first neighbours of sinks without being sinks themselves:

∞ ( υ) y2

LL

∫

=

L(yυ) (ωx + 2 y + z , t )dt = LL(yυ2) ( p )

t =0

p =0

∞

∫

LL(zυ2) =

L(zυ) (ωx + y + 2 z , t )dt = LL(zυ2) ( p )

t =0

p =0

1 VZB

∫

1 = VZB

∫

=

FLL(yυ) (k , p )

p=0

e

+ ik ωx+2 y + z

d3 k

(8) FLL(zυ) (k , p )

p=0

e

+ ik ωx+ y + 2 z

d3 k

with S y( υ2) = S z( υ2 ) . Thanks to mirror symmetry: ∞

∫

∞

L(yυ) (ω− x + 2 y + z , t )dt = − LL(yυ2)

t =0

∫

L(zυ) (ω− x + y + 2 z , t )dt = − LL(zυ2)

t =0

Unlike the quantity S y( 1υ) = S z(1υ) , the time integrals LL(yυ2) = LL(zυ2) remain finite when p → 0.

2. Transport equations 2.1 Transport equation for the function L(xυ) (r , t ) The following three lines correspond to the general term GTx(υ) describing the exchanges among the probability functions which are stem from first neighbour jumps:   dL (r , t )  = + WO  L(υy ) (r + ωx + y , t ) + L(υy ) (r + ωx − y , t ) + L(υy ) (r + ω− x + y , t ) + L(υy ) (r + ω− x − y , t )   GTx(υ) dt  +WO  L(υz ) (r + ωx + z , t ) + L(υz ) (r + ωx − z , t ) + L(υz ) (r + ω− x + z , t ) + L(υz ) (r + ω− x − z , t )   (υ) x

−8WO L(υx ) (r , t )

The right hand side must then be corrected to take into account the specificity of sinks sites: the total escape frequency from the sink sites is zero and the inward contribution of these sinks to neighbouring sinks must be subtracted:  8WO L(υx ) (ω y + z , t ) − WO L(υy ) (ωx + z , t ) − WO L(υz ) (ωx + y , t )   +δ(r − ω y + z )  (υ ) (υ )   − W L ( ω , t ) − W L ( ω , t ) O y x z O z x y − + − +    8WO L(υx ) (ω y − z , t ) − WO L(υy ) (ωx − z , t ) − WO L(υz ) (ωx + y , t )   +δ(r − ω y − z )  (υ ) (υ )   − W L ( ω , t ) − W L ( ω , t ) − − − + O y x z O z x y    8WO L(υx ) (ω− y + z , t ) − WO L(υy ) (ωx + z , t ) − WO L(υz ) (ωx − y , t )   +δ(r − ω− y + z )  (υ ) (υ )   − W L ( ω , t ) − W L ( ω , t ) − + − − O y x z O z x y    8WO L(υx ) (ω− y − z , t ) − WO L(υy ) (ωx − z , t ) − WO L(υz ) (ωx − y , t )   +δ(r − ω− y − z )  (υ ) (υ )   − W L ( ω , t ) − W L ( ω , t ) O y O z − x−z − x− y  

Then a correction is needed for the four ‘x’ sites in planes x = +2 and x = −2 , which are first neighbours of sink sites, without being sinks themselves. For all of them the contribution coming from sink sites must be deleted: +δ(r − ω2 x + y + z )  −WO L(υy ) (ωx + z , t ) − WO L(υz ) (ωx + y , t )  +δ(r − ω−2 x + y + z )  −WO L(υy ) (ω− x + z , t ) − WO L(υz ) (ω− x + y , t )  +δ(r − ω2 x − y + z )  −WO L(υy ) (ωx + z , t ) − WO L(υz ) (ωx − y , t )  +δ(r − ω−2 x − y + z )  −WO L(υy ) (ω− x + z , t ) − WO L(υz ) (ω− x − y , t )  +δ(r − ω2 x + y − z )  −WO L(υy ) (ωx − z , t ) − WO L(υz ) (ωx + y , t )  +δ(r − ω−2 x + y − z )  −WO L(υy ) (ω− x − z , t ) − WO L(υz ) (ω− x + y , t )  +δ(r − ω2 x − y − z )  −WO L(υy ) (ωx − z , t ) − WO L(υz ) (ωx − y , t )  +δ(r − ω−2 x − y − z )  −WO L(υy ) (ω− x − z , t ) − WO L(υz ) (ω− x − y , t ) 

The complete equation is Fourier and Laplace transformed according to FL(υx ) (k , t ) = ∑ e− ikr L(υx ) (r , t )

→ ∑ e − ikr L(υx ) (r + ω, t ) = FL(υx ) (k , t )eik ω

{r}

∫

∞

0

{r}

FL(υx ) (k , t )e− pt dt =FLL(υx ) (k , p ) − FL(xυ)

which yields : pFLL(υx ) (k , p ) − FL(xυ) = −8WO FLL(υx ) (k , p ) + 4WO cx c y FLL(υy ) (k , p) + 4WO cx cz FLL(υz ) (k , p)} FLGTx(υ) −e

− ik ω2 x+ y + z

(2 S y( 1υ) )WO − e

− ik ω−2 x+ y + z

(−2 S y( 1υ) )WO − e

− ik ω2 x− y + z

(2 S y( υ1 ) )WO − e

− ik ω−2 x− y + z

(−2 S y( υ1 ) )WO

−e

− ik ω2 x+ y − z

(2 S y( 1υ) )WO − e

− ik ω−2 x+ y − z

(−2 S y( υ1 ) )WO − e

− ik ω2 x− y − z

(2 S y( υ1 ) )WO − e

− ik ω−2 x− y − z

(−2 S y( υ1 ) )WO

+e

− ik ω y+ z

+e

− ik ω− y+ z

(8W S (8W S O

O

( υ) x0

− 2S y( υ1 )WO − (−2S y( 1υ) )WO ) + e

( υ) x0

− ik ω y − z

− 2S y( 1υ)WO − (−2 S y( 1υ) )WO ) + e

(8W S (8W S

( υ) x0

O

− ik ω− y − z

O

− 2 S y( 1υ)WO − (−2 S y( 1υ) )WO )

( υ) x0

− 2 S y( 1υ)WO − (−2 S y( υ1 ) )WO )

The mirror symmetry plane leads to S x( υ0 ) = 0 and to the compensation of the contributions coming from the sinks on planes x = +1 and x = −1 . The last two lines reduce to zero with the final formulation: (υ ) x

( υ) x

pFLL (k , p ) − FL

(υ ) x

= FLGT

− 2WO S

( υ) y1

 e− ik ω2 x+ y+z + e − ik ω2 x− y+ z + e− ik ω2 x+ y−z + e − ik ω2 x− y−z   − ik ω  − ik ω − ik ω − ik ω −2 x + y + z − e −2 x− y+ z − + e −2 x+ y−z − e −2 x− y−z   −e

→ pFLL(υx ) (k , p ) − FLGTx(υ) = FL(xυ) + 32icx c y cz sxWO S y( 1υ)

Hence the Laplace-Fourier transformed equation becomes:

( p + 8WO ) FLL(υx ) − 4WO cx c y FLL(υy ) − 4WO cx cz FLL(υz ) = FL(xυ) + 32iS y( υ1 )WO cx c y cz sx

(9)

2.2 Transport equations for L(υy ) (r , t ) and L(υz ) (r , t ) The function L(υx ) (r , t ) implies always a displacement of the tracer atom with a non-zero component along Ox. Functions L(υy ) (r , t ) and L(υz ) (r , t ) are different: the divacancy has a first end on the same x plane as that of the tracer atom and a second end in an adjacent x plane. When the first end jumps, the corresponding tracer jump has no component along Ox; but, as a result of this jump having no xcomponent, the correlation (i.e. the bias between the probabilities of future jumps along +x or –x) does not vanish, unlike the case of the monovacancy mechanism. Indeed, the second end of the divacancy remains the ingredient of a bias between future jumps having a positive or negative component along Ox. Let us illustrate this point with an example: the divacancy jump from the ‘z’ site at ωx + y + 2 z towards the ‘y’ site ωx + z moves the vacancy located on the FCC site ω2 y + 2 z towards the origin ; the tracer atom is displaced in the reverse direction and the resulting configuration (tracer + divacancy) is identical to the configuration (tracer on the origin + divacancy at the ‘y’ site ωx − 2 y − z ), apart from a translation of ω2 y + 2 z . This remark implies that a transition (or a fictitious ‘jump’) of the divacancy from the ‘z’ site at ωx + y + 2 z to the ‘y’ site at ωx − 2 y − z (and vice versa) must be added to account for this configurational change. The first three lines correspond to the general term for the Ly (r , t ) function; the four next to the corrections on sink sites:

  dL (r , t )  = +WO  L(υx ) (r + ωx + y , t ) + L(υx ) (r + ωx − y , t ) + L(υx ) (r + ω− x + y , t ) + L(υx ) (r + ω− x − y , t )   GTy(υ) dt  +WO  L(υz ) (r + ω y + z , t ) + L(υz ) (r + ω y − z , t ) + L(υz ) (r + ω− y + z , t ) + L(υz ) (r + ω− y − z , t )   (υ) y

−8WO L(υy ) (r , t )

+δ(r − ωx + z ) ( 8WO L(υy ) (ωx + z , t ) − WO L(υx ) (ω y + z , t ) − WO L(υx ) (ω− y + z , t ) − WO L(υz ) (ωx + y , t ) − WO L(υz ) (ωx − y , t ) ) +δ(r − ωx − z ) ( 8WO L(υy ) (ωx − z , t ) − WO L(υx ) (ω y − z , t ) − WO L(υx ) (ω− y − z , t ) − WO L(υz ) (ωx + y , t ) − WO L(υz ) (ωx − y , t ) ) +δ(r − ω− x + z ) ( 8WO L(υy ) (ω− x + z , t ) − WO L(υx ) (ω y + z , t ) − WO L(υx ) (ω− y + z , t ) − WO L(υz ) (ω− x + y , t ) − WO L(υz ) (ω− x − y , t ) ) +δ(r − ω− x − z ) ( 8WO L(υy ) (ω− x − z , t ) − WO L(υx ) (ω y − z , t ) − WO L(υx ) (ω− y − z , t ) − WO L(υz ) (ω− x + y , t ) − WO L(υz ) (ω− x − y , t ) ) At last eight lines corresponding to the four ‘y’ sites in plane x = +1 and four ‘y’ sites in plane x = −1 which are first neighbours of sinks must then be added. The terms corresponding to the extra transitions are written in red below:

+δ(r − ωx + 2 y + z )  −WO L(xυ) (ω y + z , t ) − WO L(zυ) (ωx + y , t )

+ WO L(zυ) (ωx − y − 2 z , t ) 

+δ(r − ωx + 2 y − z )  −WO L(xυ) (ω y − z , t ) − WO L(zυ) (ωx + y , t )

+ WO L(zυ) (ωx − y + 2 z , t ) 

+δ(r − ωx − 2 y + z )  −WO L(xυ) (ω− y + z , t ) − WO L(zυ) (ωx + y , t ) + WO L(zυ) (ωx + y − 2 z , t )  +δ(r − ωx − 2 y − z )  −WO L(xυ) (ω− y − z , t ) − WO L(zυ) (ωx − y , t ) + WO L(zυ) (ωx + y + 2 z , t ) 

+δ(r − ω− x + 2 y + z )  −WO L(xυ) (ω y + z , t ) − WO L(zυ) (ω− x + y , t ) + WO L(zυ) (ω− x − y − 2 z , t )  + δ(r − ω− x + 2 y − z )  −WO L(xυ) (ω y − z , t ) − WO L(zυ) (ω− x + y , t ) + WO L(zυ) (ω− x − y + 2 z , t )  +δ(r − ω− x − 2 y + z )  −WO L(xυ) (ω− y + z , t ) − WO L(zυ) (ω− x − y , t ) + WO L(zυ) (ω− x + y − 2 z , t )  +δ(r − ω− x − 2 y − z )  −WO L(xυ) (ω− y − z , t ) − WO L(zυ) (ω− x − y , t ) + WO L(zυ) (ω− x + y + 2 z , t )  Fourier-Laplace transforms lead to the final equation for FLL(yυ) : −4WO c y cx FLL(xυ) + ( p + 8WO ) FLL(yυ) − 4WO c y cz FLL(zυ) = FL(yυ) + 8iWO S y( υ1 ) cz sx (c2 y − 3) − 8iWO LL(yυ2) cz sx c2 y . (10)

Thanks to the four-fold symmetry around the Ox axis, the equation for FLL(zυ) is obtained through a mere exchange of ‘y’ and ‘z’ : −4WO cz cx FLL(xυ) − 4WO c y cz FLL(yυ) + ( p + 8WO ) FLL(zυ) = FL(zυ) + 8iWO S z(1υ) c y sx (c2 z − 3) − 8iWO LL(zυ2) c y sx c2 z . (11)

3. Calculation of return probabilities The system of Eqs. (9-11) is solved to get the explicit expression of the functions L(xυ) (r , t ), L(yυ) (r , t ), L(zυ) (r , t ) and determine the unknows S y( υ1 ) , LL(yυ2) .  FLL(xυ)   FL(xυ) + Bx S y(υ1 ) + Cx LL(yυ2)   p + 8WO   ( υ)  ( υ)  (υ ) (υ )  M  FLLy  =  FLy + By S y1 + C y LLy 2  with M =  −4WO c y cx  FLL( υ)   FL( υ) + B S (υ ) + C LL(υ )   −4W c c z  z y1 z y2  O z x   z 

−4WO cx c y p + 8WO −4WO cz c y

−4WO cx cz   −4WO c y cz  , p + 8WO 

and

Bx = 32iWO cx c y cz sx

Cx = 0

By = 8iWO cz sx (c2 y − 3) Bz = 8iWO c y sx (c2 z − 3)

C y = −8iWO cz sx c2 y . Cz = −8iWO c y sx c2 z

(12)

The general solution  FLL(xυ)   FL(xυ) + Bx S y(υ1 ) + C x LL(yυ2)   ( υ)  −1  ( υ) (υ ) (υ )   FLLy  = M  FLy + By S y1 + C y LLy 2   FLL( υ)   FL( υ) + B S (υ ) + C LL(υ )  z  z y1 z y2    z

requires the expressions of the determinant ∆ and cofactors Dij ; they are given in Appendix A. First order terms with respect to p must be kept all along. The inverse Fourier transform is used to get the time integral of the looked for probability on any desired site ‘r’ of type ‘c’:

LL(cυ) (r , p ) =

1 VZB

∫

VZB

FLL(cυ) (k , p )e+ ikr d3 k

(13)

where the superscript (υ) keeps track of the initial condition. 3.1 Determining pS y( υ1 ) et WO LL(yυ2) The system must first be solved for FLL(yυ) in order to obtain two equations for the two unknowns S y( 1υ) et LL(yυ2) :

FLL(yυ) =

N

( υ) y

∆

Ν (yυ) = ( D12

D22

 FL(xυ) + Bx S y(υ1 ) + Cx LL(yυ2)    D32 )  FL(yυ) + By S y(υ1 ) + C y LL(yυ2)   FL( υ) + B S (υ ) + C LL(υ )  z y1 z y2   z

The numerator can be recast into the form N y( υ) = N y( υ0) + N y1S y( υ1 ) + N y 2 LL(yυ2) , where the expressions of N y( υ0) , N y1 , N y 2 are established in Appendix B. Instead of using the value on the particular site ωx + z only, we perform an average on the four sink sites ωx + z , ωx − z ω− x + z ω− x − z , since the values are equal in magnitude (while taking account of the sign change induces by the mirror symmetry plane): e + ik ωx+ z + e + ik ωx−z − e+ ik ω− x+ z − e + ik ω− x−z d3 k = i ∫ FLL(yυ) sx cz d3 k 4 {N y(υ0) + N y1S y(1υ) + N y 2 LL(yυ2) } s c d k x z 3 ∆

S y( 1υ) = ∫ FLL(yυ) = i∫

Hence the first equation:   N y1 N y2 N y( υ0) S y( 1υ) 1 − i ∫ sx cz d3 k  − iLL(yυ2) ∫ s x cz d 3 k = i ∫ s x cz d3 k ∆ ∆ ∆  

It is shown in Appendix B that the coefficient of S y( 1υ) can be recast into the form pI11 WO , and that the right-hand side can be recast into the form I13( υ) WO . In this first

equation connecting the finite quantities pS y( υ1 ) and WO LL(yυ2) , the integrals I11 , I12 , I13( υ) are lattice integrals, the expression of which is given in Appendix B:

→ I11 pS y( 1υ) + I12WO LL(yυ2) = I13( υ) In the same way, we evaluate LL(yυ)2 through an average on eight sites:

(14)

LL(yυ2) = ∫ = i∫

+ ik ωx+ 2 y + z + ik ω + ik ω + ik ω + e x + 2 y − z + e x −2 y + z + e x −2 y − z e    + ik ω + ik ω + ik ω + ik ω N y( υ)  − e − x+2 y+ z − e − x+2 y−z − e − x−2 y+z − e − x−2 y−z  d3k 8 ∆ N y( υ) c2 y cz sx d3 k ∆

Hence the second equation:   N N ( υ) c2 y cz sx d3k + LL(yυ2) 1 − i ∫ y 2 c2 y cz sx d3 k  = i ∫ y 0 c2 y cz sx d3 k ∆ ∆ ∆  

N y1

−iS y( υ1 ) ∫

In the same way as above, Appendix B shows that the coefficient of S y( 1υ) and the right-hand side are of the form pI 21 WO and I 23( υ) WO respectively. The second equation can be recast into the form: ( υ) → pI 21S y( 1υ) + I 22WO LL(yυ2) = I 23

(15)

The solution of the system is immediate and is given by: pS y( υ1 ) =

( υ) I13( υ) I 22 − I 23 I12 I11 I 22 − I 21 I12

WO LL(yυ2) =

(υ) I 23 I11 − I13( υ) I 21 I11 I 22 − I 21 I12

(16)

As expected, the two unknowns being time integrals have the physical dimension of a time; the first one is proportional to 1/p; the second one is proportional to 1/ WO . The formal expressions of integrals I11 , I 22 , I 21 , I12 , I13( υ) , I 23( υ) are given in Eqs. (B1-B8); their numerical values are gathered in Table B1. It is also shown that they are not independent but rather linked by three relations established in Eqs. (B9-B11). 3.2 Calculation of the total return probabilities The calculation of the correlation factor implies the total return probability of the divacancy onto the tracer to let it perform a jump with non-zero x projection. The type of the second jump can be of any of the two jump types 1 or 2. 3.2.1 Probability of performing a second tracer jump of type 2 The total probability P∑( υy)1 of a second jump of type 2 with positive xcomponent implies the presence of the divacancy on the four ‘y’ sites ωx + 2 y + z , ωx + 2 y − z , ωx − 2 y + z , ωx − 2 y − z in the plane x = +1 ; the contribution of a jump with a negative x-component is already embedded in the antisymmetric initial condition:

P∑( υy1) = WO ∫ FLL(yυ) (e

+ ik ωx+ 2 y + z

+e

+ ik ωx+ 2 y − z

+e

+ ik ωx−2 y + z

+e

+ ik ωx−2 y − z

)d3k

= 4WO ∫ FLL(yυ) (cx + isx )c2 y cz d3 k

(17)

= Py(1υ) + Py11 pS y( 1υ) + WO Py12 LL(yυ2)

As shown in Appendix C, Py(1υ) and Py11 are dimensionless while the third coefficient can be recast into the form WO Py12 . Their analytical expressions are given in Eqs. (C1-C4) ; their numerical values are gathered in Table C1. It is also shown that Py11 , Py12 are proportional to I 21 , I 22 respectively. 3.2.2 Probability of performing a second tracer jump of type 1 The total probability P∑( υx ) of a second tracer jump of type 1 with a positive x-component implies the presence of the vacancy on any of the four ‘x’ sites ω2 x + y + z , ω2 x + y − z , ω2 x − y + z , ω2 x − y − z in the plane x = +2 ; the contribution of a jump with negative x-component is embedded in the initial condition. This probability needs the expression of FLL(xυ) (k , p ) :

FLL(xυ) =

( υ) x

N ∆

N x( υ) = ( D11 D21

 FL(xυ ) + Bx S y( 1υ)   (υ ) ( υ) ( υ)  D31 )  FLy + By S y1 + C y LLy 2   FL(υ ) + B S ( υ) + C LL( υ)  z y1 z y2   z

N x( υ) = ( D11 FL(xυ ) + D21 FL(yυ ) + D31 FL(zυ ) ) + ( D11 Bx + D21 By + D31 Bz ) S y( υ1 ) + ( D11Cx + D21C y + D31Cz ) LL(yυ2)

As above it must be Fourier inversed and the average is performed on the four sites: P∑( υx ) = 2WO ∫ FLL(xυ) (e

+ ik ω2 x+ y+ z

+e

+ ik ω2 x+ y − z

+e

+ ik ω2 x− y + z

+e

+ ik ω2 x− y − z

)d3k

= 8WO ∫ FLL(xυ) (c2 x + is2 x )c y cz d3 k

(18)

= Px( υ) + Px1 pS y( 1υ) + WO Px 2 LL(yυ2)

As above, the coefficient of S y( 1υ) is shown in Appendix D to be equal to pPx1 . The expressions of the coefficients Px( υ) , Px1 , Px 2 are established in Eqs. (D1-D4) and their numerical values gathered in Table D1. It is also shown that Px1 , Px 2 are proportionnal to I13(2) , I 23(2) respectively according to Eqs. (D6-D7). 3.3 Expressions of tij and correlation factor We calculated above the contribution P∑(1)y1 of divacancy jumps starting from any of the four ‘y’ sites of plane x = +1 . The contribution P∑(1)z1 of the divacancy jumps starting from the four ‘z’ sites on plane x = +1 is of equal magnitude thanks to the

four-fold symmetry assigning the same role to ‘y’ and ‘z’ variables. Hence the final average products tij entering the expression of the correlation factor: t11 = − P∑(1)x

t12 = −( P∑(1)y1 + P∑(1)z1 ) = −2 P∑(1)y1

t21 = − P∑(2)x

t22 = −( P∑(2)y1 + P∑(2)z1 ) = −2 P∑(2)y1

(19)

Appendix E establishes the equality t11 = t22 in Eqs. (E2-E3). The numerical values of tij’s are gathered in Table E1. The final value of the correlation factor f 2V is thus equal to: f 2V = 0.45809434

(20)

This exact value is in the confidence interval of the two best evaluations known so far.6-7

Acknowledgements We thank warmly J. Creuze from LEMHE-ICMMO, Orsay (France), with whom this work was started when he was a M2 student.

Bibliography 1

Bardeen J . and Herring C., Atom movements, J.H. Hollomon Ed. (American Society for Metals, Cleveland,1951), p. 87; Imperfections in nearly perfect crystals, W. Shockley Ed. (Wiley, NY) 1952, p. 261-288. 2 Mullen J.G., Physical Review 124 (1961), p. 1723. 3 Schottky G., Physics Letters 12 (1964) p.95. 4 Howard R.E., Phys. Rev. 144 (1966) p. 650. 5 Hehrer H. J., Phys. F: Metal Physics 2 (1978) p. L11. 6 Bennett C.H. and Alder B.J., J. Phys. Chem. Solids 32 (1971) p. 2111. 7 Murch G.E., J Phys. Chem. Solids 45 (1984) p. 451. 8 Besnoit P., Bocquet J.L., Lafore P., Acta Metallurgica 25 (1977), p. 265. 9 Bocquet J.L., Philosophical Magazine A47 (1983), p. 547 ; Philosophical Magazine A63 (1991), p. 157. 10 Bocquet J.L., http://arxiv.org/abs/1312.4390 (2013).

Appendix A: Expressions of determinant ∆ and cofactors Dij at first order in p After retaining only the terms at first order in p : p + 8WO ∆ = −4WO c y cx −4WO cz cx

−4WO cx c y

−4WO cx cz

p + 8WO −4WO cz c y

−4WO c y cz ≈ ∆ 0 + p∆1 p + 8WO

(A1)

∆ 0 = 128WO3  4 − cx2c y2 cz2 − cx2 c y2 − c 2y cz2 − cz2 cx2  = 128WO3 ∆'0 ∆1 = 16WO2 12 − cx2c y2 − c y2cz2 − cz2 cx2  = 16WO2 ∆1' ∆ = ∆ 0 (1 + p

(A2)

 ∆1 ∆1'   ) → ∆ = 128WO3 ∆'0 1 + p '   ∆0 ∆ 8 W O 0  

Alternative expressions of determinant as a function of cofactors Dij :

∆ = ( p + 8WO ) D11 − 4WO c y cx D21 − 4WO cz cx D31  D11 ≈ 64WO2 − 16WO2 c y2 cz2 + p16WO  with  D21 = 16WO2cx c y cz2 + 32WO2 cx c y + p 4WO cx c y  2 2 2  D31 = 16WO cx c y cz + 32WO cx cz + p 4WO cx cz

(A3)

or

∆ = −4WO cx c y D12 + ( p + 8WO ) D22 − 4WO c y cz D32  D12 = 16WO2 cx c y cz2 + 32WO2 cx c y + p 4WO cx c y  with  D22 ≈ 64WO2 − 16WO2cx2cz2 + p16WO  2 2 2  D32 = 16WO cx c y cz + 32WO c y cz + p 4WO c y cz

(A4)

or

∆ = −4WO cx cz D13 − 4WO c y cz D23 + ( p + 8WO ) D33  D13 = 16WO2 cx c y2 cz + 32WO2 cx cz + p 4WO cx cz  with  D23 = 16WO2 cx2 c y cz + 32WO2c y cz + p 4WO c y cz  2 2 2 2  D33 ≈ 64WO − 16WO cx c y + p16WO

Symmetry implies : D12 = D21

D23 = D32

(A5)

D31 = D13

Appendix B: Integral expressions of the coefficients I ij( υ) in the equations yielding pS y( υ1 ) and LL(yυ2)

Dij stand for the cofactors calculated in Appendix A (Eqs. A3-A5). In all what follows the normalizing factor 1/ VZB of the integrals is omitted for simplicity. B1. Calculation of

D12 Bx + D22 By + D32 Bz pI11 N11 = 1− i∫ s x cz d3 k = 1 + ∫ d3 k WO ∆ ∆

Numerator N11 :

Bx = 32iWO cx c y cz s x

By = 8iWO cz sx (c2 y − 3)

Bz = 8iWO c y sx (c2 z − 3)

N y1 = D12 Bx + D22 By + D32 Bz = −512iWO3cz sx ∆ O' + 64ipWO2 cz sx {2cx2 c y2 + c y2 cz2 + 2c y2 − 8} Multiplying by the integrating factor :

  2cx2 c y2 + c y2 cz2 + 2c y2 − 8  N11 = −isx cz N y1 = −512W c s ∆ 1 − p  8WO ∆ O'   3 2 2 O z x

' O

Hence, retaining only first order term with respect to p yields:   2cx2 c y2 + c y2 cz2 + 2c y2 − 8  4c s 1 − p  8WO ∆ O' N11   I11 = 1 + ∫ d3 k = 1 − ∫ d3k ∆  ∆1'  1 + p  '   8 W ∆ O 0     2cx2 c y2 + c y2 cz2 + 2c y2 − 8 + ∆1'  2 2 = 1 − ∫ 4cz sx 1 − p  d3k 8WO ∆ O'   c 2 c 2 − cx2 cz2 + 2c y2 + 4 2 2 2 2 x y d3k = 1 − ∫ 4cz sx d3 k + p ∫ cz sx 2WO ∆ O' 2 2 z x

The first integral is equal to 1, hence: cz2 sx2 {cx2 c 2y − cx2 cz2 + 2c 2y + 4} pI11 = p∫ d3k WO 2WO ∆ O'

(B1)

The integral S y( 1υ) for the sink sites at ωx ± y and ωx ± z appears in the first equation with the variable ‘p’ as a multiplying factor as mentioned in the main section.

B2. Calculation of I12 = −i ∫

D12C x + D22C y + D32C z ∆

s x cz d 3 k = ∫

N12 d3 k ∆

Expression of N12 : Cx = 0

C y = −8iWO cz sx c2 y

C z = −8iWO c y sx c2 z

N y 2 = D12Cx + D22C y + D32C z = −128iWO3cz sx {−cx2 c y2 + 4c y2 cz2 + cx2 cz2 + 6c y2 − 4} − p32iWO2 cz sx {4c2 y + c y2 c2 z }

The first term which is independent of ‘p’ gives a non-vanishing contribution to the integral; as a consequence, the first order term in p can be dropped:

N12 = −isx cz N y 2 = −128WO3cz2 sx2  −cx2 c y2 + 4c y2 cz2 + cx2 cz2 + 6c y2 − 4  I12 = − ∫

128WO3cz2 sx2  −cx2 c y2 + 4c y2 cz2 + cx2 cz2 + 6c y2 − 4  128WO3 ∆O'

d3 k

Hence : I12 = − ∫

cz2 sx2  −cx2 c y2 + 4c y2 cz2 + cx2 cz2 + 6c y2 − 4  ∆ O'

(B2)

d3k

D12 FL(xυ) + D22 FL(yυ) + D32 FL(zυ) I13( υ) N13( υ) s x cz d 3 k = ∫ d3k = i∫ B3. Calculation of ∆ ∆ WO

Numerator N13(1) and N13(2) : (first order terms in p can be dropped in Dij ). For CI1 : FL(1) x =0

FL(1) y = −ic2 y cz s x

FL(1) z = −ic y c2 z s x

{

}

(1) (1) 2 2 2 2    2 2  N y(1)0 = D12 FL(1) x + D22 FLy + D32 FLz = −16iWO cz s x c2 y  4 − cx cz  + c2 z  cx c y + 2c y 

{

}

N13(1) = iN y(1)0 sx cz = 16WO2 cz2 sx2 c2 y  4 − cx2 cz2  + c2 z cx2 c y2 + 2c y2 

Hence : cz2 sx2 {cx2 cz2 + 4c y2 cz2 − cx2 c y2 + 6c y2 − 4} I13(1) d3 k = 8WO ∆O' WO ∫

For CI2 : FL(2) x = −4icx c y cz s x

FL(2) y =0

(B3)

FL(2) z =0

2 2 2 2 N y(2)0 = D12 FL(2) x = −64iWO cx c y cz s x {cz + 2}

N13(2) = iN y(2)0 sx cz = 64WO2 cx2 c y2 cz2 sx2 {cz2 + 2}

Hence : cx2 c y2 cz2 sx2 {cz2 + 2} I13(2) d3k = 2WO ∆ O' WO ∫

Calculation of

(B4)

D B + D22 By + D32 Bz pI 21 N 21 = −i ∫ 12 x c2 y cz sx d3k = ∫ d3 k WO ∆ ∆

Numerator N 21 = c2 y N11 cx2 c y2 − cx2cz2 + 2c y2 + 4 pI 21 2 2 2 2 Hence : = − ∫ 4cz sx c2 y d 3k + p ∫ cz sx c2 y d3k WO 2WO ∆ O'

The first integral is equal to zero, hence: cz2 sx2 c2 y cx2c y2 − cx2 cz2 + 2c y2 + 4  pI 21 = p∫ d3k WO 2WO ∆ O'

Calculation of I 22 = 1 − i ∫

(B5)

D12Cx + D22C y + D32Cz ∆

c2 y cz sx d3k = 1 − ∫

N 22 d3k ∆

Numerator N y 2 = c2 y N12 Hence : I 22 = 1 − ∫

cz2 sx2 c2 y {−cx2 c y2 + 4c y2 cz2 + cx2 cz2 + 6c y2 − 4} ∆ O'

d3k

I 23( υ) N 23( υ) Calculation of = d3k WO ∫ ∆

Numerator N 23(1) for CI1 : N 23(1) = c2 y N13(1) = 16WO2cz2c2 y sx2 {cx2 cz2 + 4c y2cz2 − cx2 c y2 + 6c 2y − 4} (1) 16WO2 cz2 c2 y sx2 {cx2 cz2 + 4c y2 cz2 − cx2c y2 + 6c y2 − 4} I 23 = d3k WO ∫ 128WO3∆ O'

Hence :

(B6)

(1) cz2 c2 y sx2 {cx2 cz2 + 4c y2cz2 − cx2 c y2 + 6c y2 − 4} I 23 = d3k WO ∫ 8WO ∆ O'

(B7)

Numerator N 23(2) for CI2: N 23(2) = c2 y N13(2) = 64WO2cx2c y2 cz2 c2 y sx2 {cz2 + 2} (2) 64WO2 cx2 c y2 cz2c2 y sx2 {cz2 + 2} I 23 d3k = WO ∫ 128WO3 ∆ O'

Hence : cx2 c y2 cz2 c2 y sx2 {cz2 + 2} I 23(2) = d3k WO ∫ 2WO ∆ O'

(B8)

Relationships between integrals All the lattice integrals given above are not independent. We derive several useful relationships to be used later on. In order to recast the expression of an integral into an alternative formulation, we take into account the fact that the variables ‘y’ and ‘z’ play the same role and can be interchanged at will. For instance : (2) 13

I

=∫

cx2 c y2 cz2 sx2 {cz2 + 2} 2∆O'

d3k = ∫

cx2 c y2 cz2 s x2 {c y2 + 2} 2∆ O'

d3 k = ∫

cx2 c 2y cz2 sx2 {c 2y + cz2 + 4} 4∆ O'

d3k

Relationship between I12 and I13(1) An obvious result after comparing Eqs (B2-B3): I12 = −8I13(1)

(B9)

Relationship between I 21 and I12 Starting from Eq. B5 and expanding c2 y as a function of c y gives two parts : 2 I 21 = ∫

2c y2 cz2 sx2 cx2 c y2 − cx2 cz2 + 2c y2 + 4  ∆O'

d3k − ∫

cz2 sx2 cx2 c 2y − cx2 cz2 + 2c y2 + 4  ∆O'

d3k

The first part is written as half the sum of two equivalent contributions after exchanging ‘y’ and ‘z’:

2 I 21 = ∫

c y2 cz2 sx2 cx2 c y2 − cx2 cz2 + 2c y2 + 4 ∆ O'

d3 k + ∫ −∫

c y2 cz2 sx2 cx2 cz2 − cx2 c y2 + 2cz2 + 4  ∆ O' cz2 sx2 cx2 c y2 − cx2 cz2 + 2c y2 + 4 ∆O'

d3 k

Summing the two first contributions gives : 2 I 21 = ∫

4c y2 cz2 sx2 c y2 + 2  ∆ O'

−∫

cz2 sx2  cx2 c y2 − cx2 cz2 + 2c 2y + 4  ∆O'

d3 k ,

and finally: 2 I 21 = ∫

cz2 sx2  −cx2 c y2 + 4c y4 + cx2 cz2 + 6c y2 − 4  ∆ O'

Adding Eq. B2 yields: 2 I 21 + I12 = ∫

4c y2 cz2 sx2 c y2 − cz2  ∆ O'

=∫

4c y2 cz2 sx2  cz2 − c y2  ∆ O'

The right-hand side is equal to zero thanks to the symmetric roles played by ‘y’ and ‘z’, hence : 2I 21 = − I12

(B10)

Relationship between I 22 and I 21 Starting from Eqs. (B5-B6): 2 I 21 − I 22 + 1 = ∫

4c2 y c y2 cz2 sx2 {cz2 + 2} ∆ O'

d3 k

(B11)

Relationship between I 22 and I 23(1) Comparing Eqs. (B6-B7) yields: (1) 1 − I 22 = 8 I 23

(B12)

The independent integrals entering the coefficients of the linear system yielding the two unknows pS y( υ1 ) and WO LL(yυ2) can be calculated numerically with a superposition of three Gaussian quadratures along the three axes with 512 nodes on the interval [ 0, +π ] . They are gathered below in Table B1 together with the values of the unknowns for the two initial conditions.

CI1

I11 = 2.136053180 D − 01

I12 = −1.036809774D − 01

I 21 = − I12 /2

I 22 = 7.992024140D − 01

(1) 13

I

= − I12 /8

CI2 (2) 13

I

= 2.224386633D − 02

I 23(1) = (1 − I 22 )/8

I 23(2) = 1.378066830D − 02

pS y(1)1 = 7.359994075D − 02

pS y(2) 1 = 1.090708175D − 01

WO LL(1) y 2 = 2.663185317D − 02

WO LL(2) y 2 = 1.016811720D − 02

Table B1. Numerical values of integrals. The solutions corresponding to CI1 and CI2 are displayed in lines and columns 3 and 4.

Appendix C: Expression of the coefficients required for the probability of a second tracer jump of type 2

P∑( υy1) = Py(1υ) + Py11 pS y( 1υ) + WO Py12 LL(yυ2) (2) First coefficient Py(1) 1 and Py1

Py(1υ) = 4WO ∫

D12 FL(xυ) + D22 FL(yυ) + D32 FL(zυ) ∆

(cx + isx )c2 y cz d3k = ∫

Num y( υ1 ) ∆

d3 k

For CI1 :

{

}

(1) (1) (1) 2 2 2 2    2 2  D12 FL(1) x + D22 FLy + D32 FLz = N y 0 = −16iWO cz s x c2 y  4 − cx cz  + c2 z  cx c y + 2c y 

Numy(1)1 = 4WO N y(1)0 (cx + is x )c2 y cz = 64WO3c2 y cz2 sx2 {cx2 cz2 + 4c 2y cz2 − cx2 c y2 + 6c y2 − 4}

Hence : (1) y1

P

c2 y cz2 sx2 {cx2 cz2 + 4c y2 cz2 − cx2 c y2 + 6c 2y − 4}

=∫

2∆O'

d3 k

(C1)

For CI2 : (2) (2) (2) D12 FL(2) x + D22 FL y + D32 FLz = N y 0

Num y(2)1 = 4WO N y(2)0 (cx + isx )c2 y cz = 256WO3cx2 c y2 cz2 c2 y sx2 {cz2 + 2}

Hence : 2cx2 c y2 cz2 c2 y sx2 {cz2 + 2}

Py(2) 1 = ∫

∆ O'

d3 k

Second coefficient ( D12 Bx + D22 By + D32 Bz ) (c + is )c c d k = Numy11 d k Py11 = 4WO ∫ x x 2y z 3 ∫ ∆ 3 ∆ Retaining the first order term in p and keeping only the real part : Numy11 = 4WO (cx + isx )c2 y cz N y1   2cx2 c 2y + c y2 cz2 + 2c y2 − 8  = 2048W c s c ∆ 1 − p  8WO ∆ O'   4 2 2 O z x 2y

Hence :

' O

(C2)

Py11 = ∫ 16W c s c d3 k − ∫ 2 2 O z x 2y

2 2 2 2 2 '    2cx c y + c y cz + 2c y − 8 + ∆1  2c s c   d3k ∆ O'   2 2 z x 2y

The first integral is equal to zero, which yields : pPy11 = − p ∫

2c2 y cz2 sx2 cx2 c y2 − cx2 cz2 + 2c 2y + 4  ∆ O'

(C3)

d3 k

Third coefficient ( D12Cx + D22C y + D32Cz ) (c + is )c c d k = Numy12 d k WO Py12 = 4WO ∫ 2y z 3 x x ∫ ∆ 3 ∆ Only terms not depending on p can be retained: Numy12 = 4WO (cx + is x )c2 y cz N y 2 = 512WO4 cz2 c2 y sx2 {−cx2 c y2 + 4c 2y cz2 + cx2 cz2 + 6c y2 − 4}

Hence : WO Py12 = WO ∫

4c2 y cz2 sx2 {−cx2 c 2y + 4c y2 cz2 + cx2 cz2 + 6c y2 − 4} ∆ O'

(C4)

d3k

The numerical values of the coefficients are gathered in Table C1.

Py(1) 1 = 1.003987929D − 01 Py(2) 1 = 5.512267320D − 02

Py11 = −2.073619549D − 01

Table C1. Coefficients entering the definition of

Py12 = 8.031903439D − 01

P∑( υy1) for the two initial conditions CI1 and CI2.

Relationship between integrals The new integrals are not independent. Comparison of Eq. C3-C4 with Eq. B11 yields :

Py12 − 2 Py11 = 8I 21 − 4 I 22 + 4 Relationship between Py11 and I 21 :

(C5)

Py11 = − ∫ 4 I 21 = ∫

2c2 y cz2 sx2  cx2 c y2 − cx2 cz2 + 2c y2 + 4 ∆O' 2c2 y cz2 sx2 cx2 c y2 − cx2 cz2 + 2c y2 + 4  ∆O'

d3k

d3k

Comparison of Eq. C3 with Eq. B6 yields :

Py11 = −4 I 21

(C6)

At last, using Eqs. C5-C6, we get :

Py12 = 4(1 − I 22 )

(C7)

Appendix D: Expression of the coefficients entering the probability of a second tracer jump of type 1

P∑( υx ) = Px( υ) + Px1 pS y( υ1 ) + WO Px 2 LL(yυ2) First coefficient ( D11 FL(xυ ) + D21 FL(yυ ) + D31 FL(zυ ) ) Numx(υ ) ( υ) Px = 8WO ∫ d3 k (c2 x + is2 x )c y cz d3 k = ∫ ∆ ∆ The integral depends on the initial conditions CI1 or CI2 and on the expressions of the cofactors (Eqs. A3-A5). The variable p can be set to zero in all what follows. For CI1: FL(1) x =0

FL(1) y = −ic2 y cz s x

FL(1) z = −ic y c2 z s x

{

}

Numx(1) = 256WO3cx2 c y2 cz2 sx2 c2 y cz2 + 2  + c2 z c y2 + 2 

Hence: (1) x

P

=∫

2cx2 c y2 cz2 sx2  4c y2 cz2 + 3c y2 + 3cz2 − 4  ∆ O'

For CI2: FL(2) x = −4icx c y cz s x

(D1)

d3k

FL(2) y =0

FL(2) z =0

3 2 2   2 2 2 2 Numx(2) = 8WO D11 FL(2) x (c2 x + is2 x )c y cz = 1024WO  4 − c y cz  cx c y cz s x

Hence : (2) x

P

=∫

8cx2 c y2 cz2 sx2  4 − c y2 cz2  ∆O'

(D2)

d3k

Second coefficient ( D11 Bx + D21 By + D31 Bz ) Numx1 (c2 x + is2 x )c y cz d3 k = ∫ pPx1 = 8WO ∫ d3k ∆ ∆ First order terms in p are retained. In this case, the constant terms cancel each other and only the first order terms proportional to p remain in the product:

Bx = 32iWO cx c y cz s x

By = 8iWO cz sx (c2 y − 3)

Bz = 8iWO c y sx (c2 z − 3)

Numx1 = − p1024WO3cx2 c y2 cz2 sx2 {4 + c y2 + cz2 } Hence : pPx1 = − p ∫

8cx2 c y2 cz2 sx2 {4 + c y2 + cz2 } ∆ O'

d3 k

(D3)

Third coefficient ( D11Cx + D21C y + D31Cz ) Numx 2 WO Px 2 = 8WO ∫ d3k : (c2 x + is2 x )c y cz d3 k = ∫ ∆ ∆ Only terms not depending on p can be retained:

Cx = 0

C y = −8iWO cz sx c2 y

C z = −8iWO c y sx c2 z

Numx 2 = 2048WO4 cx2 c y2 cz2 sx2 {4c y2 cz2 + 3c y2 + 3cz2 − 4} Hence : WO Px 2 = WO ∫

16cx2 c y2 cz2 sx2 {4c y2 cz2 + 3c y2 + 3cz2 − 4} ∆ O'

(D4)

d3k

The numerical values of the integrals are gathered in Table D1 below.

Px(1) = 1.102453464D − 01

Px1 = −7.118037228D − 01

Px(2) = 4.236074457D − 01

Table D1. Coefficients entering the definition of

Px 2 = 8.819627712D − 01

P∑( υx ) for the two initial conditions CI1 and CI2.

Relationships between integrals Relationship between Px(1) and I 23(2) (1) x

P I

(2) 23

=∫ =∫

=∫ =∫

2cx2c y2 cz2 sx2  4c y2 cz2 + 3c y2 + 3cz2 − 4  ∆ O' c2 y cx2 c y2 cz2 sx2 {cz2 + 2}

cx2 c y2 cz2 sx2

2∆ O'

{( 2c c

2 2 y z

d3 k = ∫

d3k

(2c y2 − 1)cx2 c 2y cz2 sx2 {cz2 + 2} 2∆ O'

d3 k

}d k

− cz2 + 4c y2 − 2 ) + ( 2c y2cz2 − c y2 + 4cz2 − 2 ) 4∆

cx2 c y2 cz2 sx2 {4c y2 cz2 + 3cz2 + 3c y2 − 4} 4∆ O'

3

' O

d3k

Hence : Px(1) = 8I 23(2)

Relationship between Px1 and I13(2)

(D5)

8cx2 c y2 cz2 sx2 {4 + c y2 + cz2 }

Px1 = − ∫

∆O' cx2 c y2 cz2 sx2 {cz2 + 2}

I13(2) = ∫

2∆O'

16cx2 c y2 cz2 sx2 {2 + cz2 }

d3k = − ∫

∆ O'

d3k

d3k

Hence : Px1 = −32 I13(2)

(D6)

Relationship between Px 2 and Px(1) Px 2 = ∫

16cx2 c y2 cz2 sx2 {4c y2 cz2 + 3c y2 + 3cz2 − 4} ∆

' O

d3k = 8Px(1)

Hence : Px 2 = 8 Px(1) = 64 I 23(2) = −2 Px1

(D7)

(1) Relation ship between Py(2) 1 and Px

(2) y1

2P

=∫ =∫

=∫

4c2 y cx2 c y2cz2 sx2 {cz2 + 2} ∆ O'

2c2 y cx2 c y2cz2 sx2 {cz2 + 2} ∆O'

d3 k

d3 k + ∫

2c2 z cx2 c y2cz2 sx2 {c y2 + 2}

2cx2 c y2cz2 sx2 {4c y2cz2 + 3c y2 + 3cz2 − 4} ∆O'

∆O'

d3 k

= Px(1)

Hence : (1) 2 Py(2) 1 = Px

(D8)

Appendix E: Equality of diagonal elements t11 = t22

t11 = −8.1344937604 D − 02

t12 = −2.1305482540 D − 01

t21 = −3.5493833258 D − 01

t22 = t11

Table E1. Average products tij’s for the divacancy mechanism in the FCC lattice.

Reducing the expression of t11 Starting from expression −t11 = P∑(1)x = Px(1) + Px1 pS y(1)1 + WO Px 2 LL(1) y 2 while using Eqs. (16) and (D6-D7) yields: −t11 = Px(1) − 32 I13(2)

(1) I13(1) I 22 − I 23 I12 I (1) I − I (1) I + 64 I 23(2) 23 11 13 21 H H

(E1)

where the denominator is defined by H = I11 I 22 − I 21 I12 . Using Eqs. (B9) and (B12) transforms the numerator of the second term into − I12 / 8 and the numerator of the third term into ( I11 − H ) / 8 . Hence the final compact expression: 4 I13(2) I12 8I 23(2) ( I11 − Η) −t11 = P + + Η Η (1) x

(E2)

Reducing the expression of t22 (2) (2) Starting from −t22 = P∑(2)y1 + P∑(2)z1 = 2 P∑(2)y1 = 2 Py(2) 1 + 2 Py11 pS y1 + 2WO Py12 LLy 2 and using Eqs. (16) and (D8) yields:

I13(2) I 22 − I 23(2) I12 I (2) I − I (2) I + 2 Py12 23 11 13 21 H H (2) (2) 2 I13 ( Py11 I 22 − Py12 I 21 ) 2 I 23 ( Py12 I11 − Py11 I12 ) = Px(1) + + H H −t22 = Px(1) + 2 Py11

Using Eqs. (C6-C7) yields : −t22 = P + (1) x

2 I13(2) ( Py11 I 22 − Py12 I 21 )

H (2) (2) 8I I 8I (1 − H ) = Px(1) − 13 21 + 23 H H

Using Eq. (B10) yields finally:

+

2 I 23(2) ( Py12 I11 − Py11 I12 ) H

−t22 = Px(1) +

4 I13(2) I12 8I 23(2) (1 − H ) + H H

(E3)